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Rewrite the following expression to make it easier to read.
Fix any grammar or spelling errors.
Remove phrases that are not part of the question.
Do not remove or change LaTeX formatting.
Do not change or remove [tex] [/tex] tags.
Do not translate the question or any part of the question.
If the question is nonsense, rewrite it so that it makes sense.
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[tex]$(\sec \theta - \tan \theta)^2 = \frac{1 - \sin \theta}{1 + \sin \theta}$[/tex]
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Answer :

GinnyAnswer
Certainly! Let's solve the given equation step-by-step to show that [tex]\((\sec \theta - \tan \theta)^2 = \frac{1 - \sin \theta}{1 + \sin \theta}\)[/tex].

1. Express [tex]\(\sec \theta\)[/tex] and [tex]\(\tan \theta\)[/tex] in terms of [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex]:
[tex]\[ \sec \theta = \frac{1}{\cos \theta} \][/tex]
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \][/tex]

2. Rewrite [tex]\(\sec \theta - \tan \theta\)[/tex] using the identities:
[tex]\[ \sec \theta - \tan \theta = \frac{1}{\cos \theta} - \frac{\sin \theta}{\cos \theta} = \frac{1 - \sin \theta}{\cos \theta} \][/tex]

3. Square the expression:
[tex]\[ (\sec \theta - \tan \theta)^2 = \left( \frac{1 - \sin \theta}{\cos \theta} \right)^2 = \frac{(1 - \sin \theta)^2}{\cos^2 \theta} \][/tex]

4. Use the Pythagorean identity [tex]\(\cos^2 \theta = 1 - \sin^2 \theta\)[/tex]:
[tex]\[ (\sec \theta - \tan \theta)^2 = \frac{(1 - \sin \theta)^2}{1 - \sin^2 \theta} \][/tex]

5. Rewrite the denominator [tex]\((1 - \sin \theta)(1 + \sin \theta)\)[/tex]:
[tex]\[ (\sec \theta - \tan \theta)^2 = \frac{(1 - \sin \theta)^2}{(1 - \sin \theta)(1 + \sin \theta)} \][/tex]

6. Simplify the fraction:
[tex]\[ (\sec \theta - \tan \theta)^2 = \frac{1 - \sin \theta}{1 + \sin \theta} \][/tex]

This calculation confirms the given equation:
[tex]\[ (\sec \theta - \tan \theta)^2 = \frac{1 - \sin \theta}{1 + \sin \theta} \][/tex]

Thus, the equation holds true for all [tex]\(\theta\)[/tex] satisfying the given trigonometric identities.

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